Getting a reduced voltage from a fixed secondary voltage of a transformer

[avicenna]

TL;DR Summary

Getting a reduced voltage from a fixed secondary voltage of a transformer.

1) I have a "choke" with many copper windings on a soft iron core. I connect it to the 12V secondary of a mains transformer. I measure the voltage of the choke with a multimeter. What is the value and why?
2) I want to light up a 6V filament lamp. My supply is a 12V from a step down transformer. Can I try to tap 6V AC for the lamp by having my lamp in series with a variable sliding resistor 50Ω and gradually reducing the resistance.

 

[Baluncore]

1) I have a "choke" with many copper windings on a soft iron core. I connect it to the 12V secondary of a mains transformer. I measure the voltage of the choke with a multimeter. What is the value and why?

If you connect an inductor to a transformer, the transformer secondary voltage will appear across the inductor. Voltage value = 12 Vac. That assumes the transformer is able to provide the current drawn by the inductive choke.

I want to light up a 6V filament lamp. My supply is a 12V from a step down transformer. Can I try to tap 6V AC for the lamp by having my lamp in series with a variable sliding resistor 50Ω and gradually reducing the resistance.

If the 6 volt filament lamp has an operating resistance lower than 50 ohms, (that can be matched by the variable resistor), then you will be able to adjust the resistor, so half the voltage appears across the lamp.

[Typo fixed]

 

[avicenna]

If you connect an inductor to a transformer, the transformer secondary voltage will appear across the inductor. Voltage value = 12 Vac. That assumes the transformer is able to provide the current drawn by the inductive choke.


If the 6 volt filament lamp has an operating resistance lower than 500 ohms, (that can be matched by the variable resistor), then you will be able to adjust the resistor, so half the voltage appears across the lamp.

Thanks. Very helpful.

Isn't it true that a good choke/inductor has very high impedance? So the current drawn "I" would be very small. Primary current about I * 240/12.
NOTE: your typo error. It is 50 ohm

 

[Baluncore]

Isn't it true that a good choke/inductor has very high impedance?

Do not maximise the impedance, Z = R + j·X
A good choke has low series resistance, R, but high reactance, X.

 

[DaveE]

The impedance magnitude would depend on the frequency $|Z| = \sqrt{R^2 + (2 \pi f L)^2}$

https://www.khanacademy.org/science...-analysis-topic/ee-ac-analysis/v/ee-impedance

 

[avicenna]

The impedance magnitude would depend on the frequency $|Z| = \sqrt{R^2 + (2 \pi f L)^2}$

https://www.khanacademy.org/science...-analysis-topic/ee-ac-analysis/v/ee-impedance

Say the choke has sufficient/many copper turns and we approximate ignoring R; R= 0.
So Z = 2π * 50 * L ohm may (generally) be high. So secondary current I = 12 / Z generally low.

Primary current = I * 12/ 240 (earlier post typo as I * 240 / 12) is usually within limit of primary current rating.

 

[DaveE]

Say the choke has sufficient/many copper turns and we approximate ignoring R; R= 0.
So Z = 2π * 50 * L ohm may (generally) be high. So secondary current I = 12 / Z generally low.

Primary current = I * 12/ 240 (earlier post typo as I * 240 / 12) is usually within limit of primary current rating.

I don't know. I think you are thinking about this correctly. But your questions are completely undefined.
What is $L$, what is "high"? I've done designs where 1Ω is very low, and other designs where 1Ω is very high.
I don't know your source current ratings. You'll need to ask more specific questions I think.

 

[Baluncore]

So the current drawn "I" would be very small.

The transformer primary is designed to be inductive, sufficient to limit the magnetising flux in the transformer.

Why would you connect an inductor across a transformer?
The current that flows will be in quadrature to the transformer secondary voltage, so the power will be reactive, not real, it will be circulating energy between the power generator and your inductor. That is not a good power factor, you will upset the supply authority, and pay a penalty rate for heating their wires with; W = I²·R

 

[avicenna]

The transformer primary is designed to be inductive, sufficient to limit the magnetising flux in the transformer.

Why would you connect an inductor across a transformer?
The current that flows will be in quadrature to the transformer secondary voltage, so the power will be reactive, not real, it will be circulating energy between the power generator and your inductor. That is not a good power factor, you will upset the supply authority, and pay a penalty rate for heating their wires with; W = I²·R

My apology. I am no expert, just asking a theoretical question about putting an inductor as load to a transformer.

My questions now all cleared. I just built a crude home transformer for some purpose. I wound 100 turns AWG 19 copper wires over a 16mm upvc tube as secondary. My primary 100 turns is wound over my secondary. I put 6 long steel bolts inside the upvc tube as core. It works. Input of 1.5V gives output of 0.7V. This is what I wanted.

Thanks for your help.